Recent Homework Questions About Calculus
P(x) = R(x) - C(x) P(x) =-x^2 +400x -x^2 -40x -100 P(x) -2x^2 + 360x -100 P'(x) = -2x +360 solve for x 0 = -2x + 360
Saturday, December 7, 2013 at 2:43pm
Calculus (check my work)
Yes! Thank you! I do have another question, though. The second part of the problem says to "compare R(11) -R(10) with that result." Does that mean that I derive it before plugging in 11 and 10 or...?
Saturday, December 7, 2013 at 2:04pm
Find the maximum revenue using the following equations: R(x)=-x^2 +400x and C(x)=x^2+40x+100. What I've done so far is use R(x) and solve for x, which got me 0 and 400. Is that how you're supposed to start it? What happens next? (I apologize; I'm really bad at ...
Saturday, December 7, 2013 at 1:38pm
Ohhhh. I see what I did wrong in that second one! Thanks for all your help!(:
Saturday, December 7, 2013 at 12:31pm
if you can find the critical numbers, that is where the derivative is zero. That is the point of min/max of the function. Check some of the related links below; you will find such problems worked out. Also, a net search for minimum revenue and such will produce many examples ...
Saturday, December 7, 2013 at 11:39am
f undefined for x=0 f' = 2-6/x^2 f'=0 when x=±√3 I'll let you sort them out.
Saturday, December 7, 2013 at 12:36am
The function crosses the origin so I see it as A = ∫ x(x^2+16)^(1/2) dx from 0 to 3 = [ (1/3)(x^2+16)^(3/2) ] from 0 to 3 = (1/3)(25)^(3/2) - 0 = (1/3)(125) = 125/3
Saturday, December 7, 2013 at 12:07am
Solve the equation 2x + 5y - 3z = -1. Write the general solution as a matrix equation. I did some guessing and ended up getting this much: [x]= [-1/2 _ _ ] X  [y]= [ 0 1 0 ] X [s] [z]= [ 0 0 1 ] X [t] however I still can't figure out the last two values in the first row
Friday, December 6, 2013 at 7:39pm
I still don't really understand how to do it... I did some guessing and ended up getting this much: [x] [-1/2 _ _]  [y]= [0 1 0] X [s] [z] [0 0 1] [t] however I still can't figure out the last two values in the first row.
Friday, December 6, 2013 at 1:28am
Calculus- please help
i cant help
Sunday, March 4, 2012 at 12:27am